Mathematics

mdave16

00:00

maximum races in an efficient algorithm? (would that suit better?)

Kevin Driscoll

I would phrase it int erms of an algorithm., ya

Daminark

00:27

I mean ofc the point is that you're looking for the fewest number of races that would guarantee you've found the fastest one, which means algorithm

gian

Is direct limit the same as colimit?

MatheiBoulomenos

01:15

@gian every direct limit is a colimit, but not the other way around. Direct limits are restricted in terms of their index category in comparision to general colimits

gian

Restricted in what way?

MatheiBoulomenos

the index category needs to be a directed set

gian

So... a poset?

MatheiBoulomenos

A special kind of poset. Do you know what a filtered category is?

gian

No.

Kasmir Khaan

01:19

Hello mathei :D

MatheiBoulomenos

Okay, well, a poset $X$ is called a directed set iff for any $x, y \in X$ there exists a $z \in X$ with $x \leq z$ and $y \leq z$

@KasmirKhaan hello!

Kasmir Khaan

@MatheiBoulomenos in the defintion of ring homomorphism, we have that f (a+b) = f(a) +f(b) , so this is a group homomprphism on the set R with addition, but we only have close on (R,*) with multiplication, my question is why do we have f(ab) = f(a) f(b) =p

I mean we dont have a group on (R,*)

gian

So essentially every subset just needs to have an upper bound.

That makes sense.

MatheiBoulomenos

@gian yes, it is equivalent to the fact that every finite subset has an upper bound

@KasmirKhaan the reason is that multiplication is part of the ring structure, or to phrase it loosely, it is part of what makes a ring interesting as a ring. If we have a homomorphism between rings, we want it to "be compatible" with this structure. If we would only consider homomorphisms of the underlying additive groups, then homomorphisms would tell us nothing about multiplication.

Kasmir Khaan

aha nice =p

I think i mixed some ideas up ><

I thought homomorphism of (R,.) had to force (R,.) to be a group

but that is clearly not the case

Leaky Nun

01:39

@MatheiBoulomenos is there any difference between monomorphism and injective homomorphism, and between epimorphism and surjective homomorphism?

Kasmir Khaan

@LeakyNun no leaky they are the same thing

MatheiBoulomenos

@LeakyNun well, for one thing, we can actually define monomorphisms and epimorphisms in every category

Kasmir Khaan

-.-

am very sure they are just formal names

Leaky Nun

@MatheiBoulomenos and then?

MatheiBoulomenos

@LeakyNun if you have a concrete category, every surjective morphism is epi and every injective morphism is mono, but there are counterexamples for each of the other directions

Leaky Nun

01:41

@MatheiBoulomenos could you give me one?

MatheiBoulomenos

@LeakyNun consider the inclusion $\Bbb Z \to \Bbb Q$ in the category of rings. This is actually an epimorphism in the category of rings. I'll leave it as an exercise for you to show this

Leaky Nun

@MatheiBoulomenos but what is an epimorphism?

MatheiBoulomenos

@LeakyNun a morphism $A \xrightarrow{f} B$ is called an epimorphism iff for every object $C$ and pair of morphisms $B \xrightarrow{g,g'} C$, $g \circ f = g' \circ f$ implies $g=g'$

Leaky Nun

oh lol

MatheiBoulomenos

@LeakyNun where $\circ$ means the composition operation which is part of the category, for concrete categories, this is function composition

@Leaky Maybe the first thing you should do is to convince yourself that epimorphisms in the category of sets with mappings as morphisms and the usual function composition are precisely surjections

Leaky Nun

01:55

Let $r \in \Bbb Q$, and $\Bbb Q \xrightarrow{g,h} R$. Then, let $r = \dfrac p q$, where $p, q \in \Bbb Z$. Then, $qg(r)=qh(r)$ with external multiplication (!). Then, $q(g(r)-h(r))=0$. It remains to prove that, well, how do I prove that $g(r)-h(r)=0$? I mean, $R$ could be something like $\Bbb Z_6$

MatheiBoulomenos

there's no homomorphism from $\Bbb Q \to \Bbb Z_6$

Leaky Nun

How do I prove $g(r)-h(r)=0$ from $q(g(r)-h(r))=0$?

@MatheiBoulomenos

MatheiBoulomenos

I'm not sure if taking differences is going to help you much

I should note that I require every ring homomorphism to take $1$ to $1$

Every rational number can be written as $n \cdot \frac{1}{m}$

for $n$ and $m$ integers

Leaky Nun

@MatheiBoulomenos sure, we require $mg(1/m) = mh(1/m) = 1$

oh, lol

that was quite non-trivial

wait, my logic was wrong

I just confused myself by treating external product as internal product

MatheiBoulomenos

External multiplication confusing for this problem, it obfuscates what's going on

Leaky Nun

02:07

I should now write $\underbrace {g\left(\frac1m\right) + g\left(\frac1m\right) + \cdots + g\left(\frac1m\right)} _ {m~\text{terms}} = \underbrace {h\left(\frac1m\right) + h\left(\frac1m\right) + \cdots + h\left(\frac1m\right)} _ {m~\text{terms}} = 1$

I must be missing something obvious

MatheiBoulomenos

I think you're confused by the fact that I chose $\Bbb Z$ as an example, it makes you think too much about addition. The same argument will actually show that $\Bbb R[x] \to \Bbb R (x)$ is also a ring epimorphism.

Leaky Nun

the thing is I have no idea what $R$ looks like

MatheiBoulomenos

You have $g(\frac{1}{m}) \cdot g(m) = h(\frac{1}{m}) \cdot h(m)$

Leaky Nun

oh nvm lmao

MatheiBoulomenos

@LeakyNun think about units

Leaky Nun

02:14

@MatheiBoulomenos ya I solved it

MatheiBoulomenos

Okay

Leaky Nun

What does it mean that $\array{ x & \xrightarrow f & y \\ f \downarrow & & \downarrow \operatorname{Id} \\ y & \xrightarrow {\operatorname{Id}} & y}$ is a pushout diagram? @MatheiBoulomenos

that took me 100 years to type lol

MatheiBoulomenos

@LeakyNun I'm sorry, I'm too lazy to TeX that kind of diagrams right now

Leaky Nun

x --f--> y

|        |
f        id
|        |
v        v

y --id-> y

MatheiBoulomenos

Thanks, but I'd need diagonal arrows to explain what a pushout diagram is

Leaky Nun

02:18

eh ok

MatheiBoulomenos

If you have too much time and want to learn this kind of stuff, I'd recommend these videos: youtube.com/watch?v=yeQcmxM2e5I&list=PLE337D7DEA972E632

Leaky Nun

x
 \
  \
   v
    y

Semiclassical

like this one: upload.wikimedia.org/wikipedia/commons/thumb/3/3b/…

?

Leaky Nun

@MatheiBoulomenos is this good?

MatheiBoulomenos

@LeakyNun ncatlab is not intended for beginners in category theory

Leaky Nun

02:21

oh ok

Semiclassical

ncatlab is not intended for beginners ~~in category theory~~

ncatlab is not intended for ~~beginners in category theory~~ anyone

5

Leaky Nun

so the terminal object of Grp is the trivial group?

MatheiBoulomenos

yes

Leaky Nun

what is the terminal object of Field?

MatheiBoulomenos

there is none

Leaky Nun

02:23

good

the terminal object of the category of vector spaces must be the dot?

the category of fields of char 2 has terminal object $\Bbb F_2$ right

MatheiBoulomenos

No

Well, yes, to your question about vector spaces, but no to your question about the category of fields of char 2

Leaky Nun

right, because, alright

MatheiBoulomenos

Though I prefer to say zero vector space rather than dot, as I prefer "A Nongeometric Approach" to things

Leaky Nun

@TedShifrin

@MatheiBoulomenos do you have any example with fields?

so a terminal object's Aut group must be trivial

MatheiBoulomenos

@LeakyNun fields are really strange as a category. But for any $p$ where $p$ is prime or $0$, there's an initial object in the category of fields of that characteristic

Leaky Nun

02:32

(is Aut group defined for categories?)

MatheiBoulomenos

Yes, Aut group is defined for categories

Leaky Nun

1 min ago, by Leaky Nun

so a terminal object's Aut group must be trivial

is this right?

MatheiBoulomenos

Yes, this is correct

Leaky Nun

hmm, interesting

well the category of $\Bbb F_2$ itself certainly has a terminal object?

MatheiBoulomenos

yes, if you take a one-object category with just one morphism from that one object to itself, you have a terminal object

Leaky Nun

02:39

can there be... no morphism?

MatheiBoulomenos

no, you gotta have an identity

Leaky Nun

this is really interesting lol

MatheiBoulomenos

well, if you have no objects, then you have no morphisms

the empty category is a thing

Leaky Nun

@MatheiBoulomenos that means every morphism from $\Bbb F_p$ to other fields of char $p$ must be unique

MatheiBoulomenos

@LeakyNun it is

and also that there always exists one

Leaky Nun

02:43

lol I got confused, I thought initial objects would be very big

somehow initial and terminal are both small

MatheiBoulomenos

@LeakyNun terminal objects can be small or big depending on the context

Leaky Nun

category of category @_@

Leaky Nun

02:58

@MatheiBoulomenos bist du doch hier?

MatheiBoulomenos

ja

Leaky Nun

do fields have product?

can I say that $\Bbb F_3[i]$ is the product of $\Bbb F_3$ with itself?

MatheiBoulomenos

@LeakyNun no. As I said, fields are really strange as a category, because every morphism is injective

Leaky Nun

@MatheiBoulomenos :o

@MatheiBoulomenos let's say I define a category on $\Bbb Z$ by saying that $a$ has a morphism to $b$ iff $a \le b$?

is this valid?

MatheiBoulomenos

@LeakyNun yes, this works

In fact, it works for any poset

Leaky Nun

03:05

wat

MatheiBoulomenos

yeah, every poset is a category

Leaky Nun

what if I add another morphism for each $a<b$

so there are two morphisms from $3$ to $5$

it's really exciting

more exciting than Galois theory

I'm just laughing with excitement

Daminark

03:18

Hai

Leaky Nun

hi

Daminark

How's it going?

MatheiBoulomenos

Hi @Daminark

Daminark

03:38

Hey @MatheiBoulomenos!

Oh I've been meaning to ask actually, do you know of a good intro to ANT?

I'm betraying topology don't tell Balarka

2

Secret

06:15

Last night dream, there's one scene where a villain is killed by taking ln of him twice, thus giving the equation: $\ln \ln x = \ln \ln 1$

user21820

06:29

@Semiclassical Sounds quite hard. It is easy to get upper bounds, but hard to prove that you can't do better. For example, a simple solution would be to race 5, then take the top 3 and race them against another 2, and repeat until done. You need 11 races this way.

A better solution is to race them in groups of 5, then race the fastest of each. That takes 6 races but is not enough because the fastest 3 need not be in different groups. But now you have 5 groups x[1..5] and y[1..5] and z[1..5] and u[1..5] and v[1..5] and x[1] > y[1] > z[1] > u[1] > v[1]. The best is x[1] and the next best 2 must be in x[2..3]+y[1..2]+z[1], so race them. This takes 7 races in total.

I don't immediately see how to prove that you can't do it in 6 races.

@Daminark @KevinDriscoll @mdave16: See above ^^ =)

user21820

06:46

Oh I know. Each race creates directed edges between those it compares. We can reduce to the transitive reduction. Then the final graph must be connected otherwise some horse isn't raced. But each race only creates 4 edges, so 6 races would only have 24, so the final graph must be a tree. Thus from the first group X only 1 horse can be re-raced. If the 1st in X was re-raced, then the 2nd in X could be in the top 3 but we can't tell. If the 1st in X wasn't re-raced, then we can't find the best.

My proof does not apply if you just want the top 3 horses in any order; though I'm pretty sure 6 races still cannot do it.

John Doe

06:57

Oi quick question

Is an inverse element only defined if we got a two-sided identity element or does it make sense to speak of it if the identity element is left or right sided?

Secret

There exists one sided inverses and one sided identities, such as left or right inverses or identities, but they only appeared in structures weaker than rings and groups

In particular, one sided inverses and identities are not necessary unique

One common example is a function that is surjective and a function that is injective. An injective function has a left inverse, while a surjective one has a right inverse

user21820

@Secret Not quite true. Even if you 'weaken' the group axioms to only assert existence of left-identities and left-inverses, it turns out that it is equivalent to the original axioms since you can prove the existence of right-identities and right-inverses! It's a curious exercise/puzzle.

Secret

ah yes, I forgot the interdependence of a certain combination of one sided identities and inverses

yes, for a certain combination of one sided inverses and identities (such as the one user21860 pointed out), you can prove they are unique and two sided

14

Q: Is a semigroup $G$ with left identity and right inverses a group?

Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group? Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a ...

you will need combinations like left identity and right inverses to be not a group

user21820

@JohnDoe Namely, left-identity existence in a group G with binary operation · and identity e is the axiom ∀x∈G ( e·x = x ), and left-inverse existence is the axiom ∀x∈G ( ∃y∈G ( y·x = e ) ).

Stupid chat likes to make asterisks into italics.

John Doe

Hmm

The thing is I only see the definition of an inverse element being made on the basis of an identity, never a left or right identity, so I'm not sure if strictly mathematically speaking, is there any sense to speak of inverses of an identity that is not two-sided.

But it may be that the definitions we are using are just not considering those or may be including them within the notion of "identity element"

user21820

07:14

@JohnDoe Well; whether the axioms make sense or not depends completely on whether the structure you intend to capture by them makes sense or not.

Secret

One sided identities are not very common outside of most maths we are doing, because semigroups are very weak structures

Kirill

The relation on $\mathbb{R}$ is defined as: $xRy$, wenn $x-y \in \mathbb{Q}$. I show that the thing is transitive: Let $a,b,c \in \mathbb{R}$ and $aRb, bRc$. $aRb$ means that $a-b \in \mathbb{Q}$. $bRc$ means that $b-c \in \mathbb{Q}$. As the sum of two rational numbers is a rational number, $(a-b)+(b-c) = a - c$ is also an element of $\mathbb{Q}$. But that means, that $aRc$ and the relation is transitive.
Do you like the argumentation?

user21820

@Kirill I'll not use "let a,b,c ..." but rather "given/take any a,b,c ..." Other than that it's a fine argument.

Kirill

@user21820 yes, I have a point here. The formal definition requiers the transitivity to work for all elements of the given set. At the same time I cannot work with all elements of the set, but pick 3 random elements.

user21820

That is why I purposely choose different words. I do not like the use of "let" in many textbooks for two completely different meanings.

When you have a real number x, you find nothing wrong with saying "let y = 3x^2−4", which uses "let" to mean that you actually name an existing object (given by a previously deduced existential statement). So I see no reason to overload "let" to also mean something else when English doesn't even support it.

Balarka Sen

07:23

@Daminark it goes it goes it goes it goes GULAG

Death Grips - Guillotine (It goes Yah)

►

user21820

@Kirill: Do you get my point? If not, feel free to clarify further.

Secret

4.00x faster

Kirill

@user21820 not really. Every proof in German starts with something like "Let $a \in \mathbb{R}$", what actually means "prepare your workspace and put a real number on it you going to operate on". So, is your point about tht the usage in English, or about the logic in general?

Secret

[Random]

$ ... \in a \in b \in c \in d \in e \in ...$

hence it is not well founded

[Ultrarandom]

user21820

@Kirill It does sound from your gloss that the general notion is flawed (even though I don't know German). Mathematicians all know how to tell what the true logical structure is, but most students do not. How about we continue in the Logic room?

♦ $Mathematics$ Logic

This room is meant for discussion about logic, including found...

1

Secret

07:38

[end ultrarandom]

Kirill

@user21820 if such cellular compartment will be usefull, we can surely do that. At the same time, my time set is bounded, so I will not have too much time to stay in chat :)

Secret

[ultrarandom]

in The h Bar, 38 secs ago, by Secret

Construct a word salad algebra with the following properties: There exists a dedekind sunset, its homorphic inverse is given by the pi cateogory of topos such that it is a cofinite and cofinal functor, with initial segment which satisfy the model of a word salad logic. Now, inject the set of all epidermal reals into the sun. This should result in a corruption of the underlying formal language, such that an abstract nonsenss is produced. Now take the uncountable orbits of $🌀^🌀(🌀)$

NB: I obviously knew almost nothing about 99% of the terms I used here, and the word "word salad" should make it clear

Had I not used the word "word salad" I will be convicted the crime of deliberately pretentious

14 hours ago, by ÍgjøgnumMeg

Hi chat

13 hours ago, by Ted Shifrin

I usually get serial downvotes, but it hasn't happened in a while.

5 hours ago, by Semiclassical

ncatlab is not intended for ~~beginners in category theory~~ anyone

5 hours ago, by Leaky Nun

this is really interesting lol

5 hours ago, by MatheiBoulomenos

yeah, every poset is a category

14 hours ago, by Liad

i think i see the idea. not sure how to formulate it

5 hours ago, by MatheiBoulomenos

Though I prefer to say zero vector space rather than dot, as I prefer "A Nongeometric Approach" to things

4 hours ago, by Daminark

I'm betraying topology don't tell Balarka

9 hours ago, by Semiclassical

“You, a very wealthy aristocrat, own 25 horses. You’re bored one day (you’re bored everyday) and decide to amuse yourself by identifying the three fastest horses in your stable.However, your personal racetrack is severely lacking in capacity, and you can only race five horses at a time. What is the minimum number of races you’ll need to organize to identify your three fastest horses?”

Oct 24 at 17:43, by Semiclassical

Not everything you find annoying needs moderator intervention

gah, they don't really fit that nicely into a conversation

Thought I can create a new timeline where this is an actual event

user21820

@Secret You just did. A nonsensical series of messages, I must say. =P

Secret

Well, 4 of the messages can actually form a coherent conversation starting with semiclassical talking about ncat and Liad said not sure how to formulate it

but anything else don't quite fit in

But anyway, for me, time is not a very continuous concept to me. I can well go back in time to murder countable many of my past selves, and then use some really advanced technology to redirect causality so that it does not erase me in the present from existence

This is assuming time travel works like that in back to the future and ripple effects actually behave like waves so you can do some notion of optics on them

You see, an interesting thing about a bunch of messages is that without the timestamps, there can exists a collection of them such that they are actually in achoronological order, but they make sense and flow nicely

I am not very sure whether it is because of the context sensitive nature of each message, or something else that provide sticky ends to thread them together

user21820

08:00

@Secret It's called finding patterns where there are none. Many humans succumb to it. =P

Secret

well we are all humans and our meaty brain is kinda limited

Also crime movies (and sometimes some real crimes as well) shows this kinda of pattern finding can forge a scene with a false story, and only very careful investigators will find the flaw between each piece

Which then brought a more philosophical topic:

How do a series of events knew they are in choronological order in time, given there are many ways to get an apparent pattern?

user21820

@Secret Events don't know anything...

Secret

oops...

Perhaps I should have ask: Given a series of events which there are many arrangements to give an apparent pattern, how do we knew which arrangement is (are) the true pattern(s)

user21820

@Secret We don't, unless there are known causal relations? We can guess though.

Grammar check: *how do we know. *should have asked

Secret

I guess that's why crime investigation is such a hard job

user21820

08:11

Lol.

Secret

(yeah: I am from Hong Kong like Leaky, and chinese languages are tenseless. This is why we made such mistakes a lot)

user21820

Oh I see.

I've always heard that Chinese does have tense, but definitely not in the morphological form of the verbs.

Secret

In chinese, the notion of tense is by juxaposition of time related nouns with the subject, that is, it is often inferred from context

user21820

Yea.

@Secret: Briefly, "do[es] <verb>" must have verb being the infinitive (bare form). "should have <verb>" must have the verb being the past form.

Secret

yup

user21820

08:18

Questions require the helper verb and puts it first, so question form of "we know" is constructed like this: "how we know" > "how we do know?" > "how do we know?". Same for question form of "we will know": "how we will know?" > "how will we know?"

Anyway many natural languages have more exceptions than rules. Sometimes annoying.

Secret

O, I did not knew that swapping the order of words makes a difference in English

user21820

Well the intermediate steps aren't valid English, but seem to explain the construction well.

English is very particular about word order.

(Wrong) Is English particular very about order word.

Secret

09:51

You can postulate that bounded but divergent limits have a value taken from some algebraic structure, but in building such structure, you need to convince the community what we can gain from being able to manipulate and assign values to these limits (how many nice rules breaks does not concerns me, as long the gain justify its clause). Meanwhile, I don't knew of any literature that tries to assign a value to these limits — Secret 9 secs ago

I am the type of person who don't worry about breaking anything just to fullfill my curiosity.

Having said that, I tend to be a bit more conservative when the consequences is dire or can potentially harm me

Secret

10:33

> This sentence is false

It is a fact that this sentence is false

It is not a fact that this sentence is false

I don't see what's the problem with "This sentence is false" having a well defined truth value (that cannot be proven) in the same way as "This is an apple"

as if "false" is a property like "apple"

But I think my mind is a bit crazy today...

Ok I think I knew what I am thinking along the lines of: I am treating "this sentence is false" as if it is one object, and not a sentence negating itself. In such view, then false becomes immutable under operations of negations

Unexpected hanging paradox

The unexpected hanging paradox or hangman paradox is a paradox about a person's expectations about the timing of a future event that he or she is told will occur at an unexpected time. The paradox is variously applied to a prisoner's hanging, or a surprise school test. Despite significant academic interest, there is no consensus on its precise nature and consequently a final correct resolution has not yet been established. One approach, logical analysis, suggests that the problem arises in a self-contradictory self-referencing statement at the heart of the judge's sentence. Epistemological studies...

O and this is a cool one:

as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on Friday.

the problem is that the prisoner have no way to tell whether the bolded statement is true or false

reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday noon, the hanging must occur on Thursday

and thus without knowing the truth value of Thursday (which he can only knew when today is thursday) he cannot have eliminated friday, and hence the whole induction like argument falls apart

Sorites paradox

The sorites paradox (; sometimes known as the paradox of the heap) is a paradox that arises from vague predicates. A typical formulation involves a heap of sand, from which grains are individually removed. Under the assumption that removing a single grain does not turn a heap into a non-heap, the paradox is to consider what happens when the process is repeated enough times: is a single remaining grain still a heap? If not, when did it change from a heap to a non-heap? == The original formulation and variations == === Paradox of the heap === The word "sorites" derives from the Greek word for heap...

and this one showed the case where the inductive step should have failed somewhere

for if it doesn't, then given those premise, then you must accept that no grains of sand will make a heap and this is valid given the premises and no extra context

It does seemed that pondering about infinite sets does really help on understanding what induction means

My solution is the same as the hysteresis solution, since it is most closely resembles how actual people will behave when doing this experiment

MatheiBoulomenos

11:27

@Daminark Milne's notes are pretty good and they're free: jmilne.org/math/CourseNotes/ANT.pdf

mdave16

@Daminark Milne is quite good, but other free online notes are available here. But if you're purely after an introduction, most of them are the same. If you are after a specific area of Alg Num Theory, then I could give you a better recommendation

ÍgjøgnumMeg

11:47

@MatheiBoulomenos Love milne's notes

robjohn

12:20

@Secret If that were only true (rather than true and painful) If only that were true (but it isn't)

Kirill

12:59

Hey guys, how would you understand $q \in 2$?

Alessandro Codenotti

it probably means that $q$ is either $0$ or $1$, but it might depend on the context

Kirill

context: $\bigcup_{q \in 2}A_q = \mathbb{R}$

ÍgjøgnumMeg

@Kirill I mean.. you can define $2 = \left\lbrace 0, 1\right\rbrace$, so maybe it just means $q \in \left\lbrace 0, 1\right\rbrace $

Kirill

where $A = \{a \mod 1 \mid a \in \overline{A} \}$, and $\overline{A}$ is the set containing representatives of each equivalence class defined by a relation $R$. Sorry for too much letters.

you define a relation on $\mathbb{R}: xRy $ if $x-y \in \mathbb{Q}$.

and $A_q = \{ x + q \mid x \in A\}, q$ rational

so, the union of all such $A_i$ shoud give me the whole $\mathbb{R}$.

but then, $q$ should be able to take any value in $\mathbb{R}$, and in combination with $q \in 2$ it sounds confusing.

So, you can also write $A$ as $[0,1]$ I think, but in context

I am not sure, but I think I know what it means

that is my Prof writing $\mathbb{Q}$ in the manner it looks like $2$!

Earth Cracks

14:02

Is there a name for a decomposition of a matrix into a product of upper triangulars, and then a permutation matrix, and then a product of upper triangulars?

(For any matrix in $\text{GL}(n,\Bbb C)$ say)

gian

14:17

Can someone help me understand why the stalk of a presheaf at a point $x$ is the same as the direct limit of open sets containing $x$?

MatheiBoulomenos

@gian isn't this the definition of a stalk?

gian

Right, I just realized. So my problem is actually computing the direct limit to arrive at the disjoint union modulo the equivalence relation.

Rajesh Dachiraju

Anone here interested in machineLearning?

HyperNeutrino

14:38

@Secret I don't get why your message got flags on it, but regardless, please don't post an unnecessary amount of chat oneboxes in chat; they can clutter up the room. I'm not sure about the exact policies of this chat room yet, so if that's acceptable, great. FYI I counterflagged it.

gian

@MatheiBoulomenos, am I right to be thinking about the cocones whose apices are groups of sections contained in the other "base" groups?

Never mind. I got it now.

user193319

If $R$ is a commutative ring, why is $(a) \subseteq (ab)$, where $(a) = \{ra + na \mid r \in R, n \in \Bbb{Z} \}$ is the ideal generated by $a$? If $R$ were a division ring, it would be trivial...but $R$ isn't assumed to be a division ring.

Alessandro Codenotti

@Daminark you remember that exercise we discussed a while back about writing elements of $S_n$ as product of two involutions? Turns out there was the same exercise in a group theory pset I was supposed to do $3$ weeks ago but I didn't :P

Your definition is strange, shouldn't $(a)$ just be $\{ra|r\in R\}$? @user193319

user193319

No. At least it hasn't been shown equivalent to that yet. What I have is the following: if $a$ is in the center of $R$, then $(a) = \{ra + na \mid r \in R, n \in \Bbb{Z} \}$.

Here's another source of confusion: how could $(a)$ contain $a$ if $R$ is not a unital ring?

Here's what I am working on: If $P$ is a prime ideal and $R$ is a commutative ring, then for every $a,b \in R$ with $ab \in P$, it follows $a \in P$ or $b \in P$.

Alessandro Codenotti

14:59

You said your ring is commutative earlier, being in the center doesn't seem very restrictive :P

user193319

Here's what the author write: "If $R$ is commutative, this implies that $(a)(b) \subseteq (ab) \subseteq$, whence $(a)(b) \subseteq P$. If $P$ is prime, then either $(a) \subseteq P$ or $(b) \subseteq P$, whence $a \in P$ or $b \in P$."

Alessandro Codenotti

what's the source?

user193319

@AlessandroCodenotti Hungerford.

I agree that $(a)(b) \subseteq (ab)$, and therefore $(a)(b) \subseteq P$. I also agree that this implies $(a) \subseteq P$ or $(b) \subseteq P$. But I don't see how we can conclude from this that either $a \in P$ or $b \in P$ since $R$ isn't assumed to be unital.

user193319

15:15

0

Q: Necessary Condition for Prime Ideal in Commutative Ring

If $P$ is an ideal in a ring $R$ such that and (1) for all $a,b \in R$ then $P$ is prime. Conversely if $P$ is prime and R is commutative, then P satisfies condition (1) The forward implication was easy to prove. However, I am having trouble with the converse, and Hungerford's explanation do...

abstract-algebra ring-theory ideals prime-ideals

I asked it on MSE main.

@AlessandroCodenotti Also the reason why $(a)$ is $\{ra + na \mid r \in R, n \in \Bbb{Z} \}$ rather than what you are familiar with is that $R$ isn't necessarily unital so that factorizations like $ra + na = (r+ n1_R)a$ cannot occur.

Faust

15:42

too lazy

thought it surprising theres something i can actually understand

alos u used wierd notation for generators

Daminark

Hai

Faust

Morning @Daminark4

Daminark

How's it going?

Faust

not great

lotsa hw really stressed out didnt do much of anything yesterday

Daminark

Oof

Well at least you've got the weekend to hopefully catch up

Faust

15:49

yeah if i can find the motivation to do some hw i think ill be ok

family stuff sucks

i got a midterm on the 8th and four assinments due this week

Daminark

Good luck

Alessandro Codenotti

hi @Dami

Daminark

How's it going @Alessandro?

Alessandro Codenotti

studying for a group theory exam

could be worse :P

Did you see the message I pinged you with earlier?

Daminark

True that. Good luck to you too

Oh lemme check

presses F to pay respects

Alessandro Codenotti

15:54

here

Daminark

I imagine they won't take it late?

Alessandro Codenotti

oh, no, we're not supposed to hand it in

the professor publishes weekly psets on the course's site, it's up to us whether to do them or not

Daminark

Ah, I see

Here psets form some part of our grade, depending on the professor it can range from 10% to 50%

Is this a general website people can see or is it password protected?

Alessandro Codenotti

Everyone can see it, but the problems are in Italian

Daminark

Ah, lmao

Alessandro Codenotti

16:06

The professor is actually a very active user on MSE

Daminark

Oh that's actually pretty nifty. I don't know too many people here who are active on MSE

Emerton used to post some answers but it's been a long time now

Faust

most of my profs dont do much here mostly on MO now

Daminark

The same guy hasn't been on MO in years

One of my profs posts every now and then on MO

Alessandro Codenotti

The algebra one is the only among my professors who is regularly active here

A couple more posted a few answers years ago

Faust

eh i goto a wierd university with alot of wierd math profs

eems budney hase been very active since he got married

the wierd part of that statement is definatly that he got married

Daminark

16:29

L M A O

Faust

well he the before that that i talked with him he was living with some 40 yr old guy downtown who rode people around the city on bikes 3 months a year for a living

The guys a fking genius but hes kind of odd and that says alot cause i have autism...

Leaky Nun

16:59

@AlessandroCodenotti posets or problem sets? :P

Daminark

I love me a weekly poset

Also yo @Eric and @Mathei

MatheiBoulomenos

yo

Daminark

How's it going?

MatheiBoulomenos

Pretty good. And for you? Pretty cool that you're interested in ANT

It's the coolest thing I've learned so far

Daminark

Yeah, I've had a lot of fun with algebra and number theory, so I do want to try this out. Though it seems like you want to know Galois theory before number theory

MatheiBoulomenos

17:11

You definitely want to know Galois theory

Leaky Nun

@MatheiBoulomenos :D

i can't decide whether Galois theory or category theory is more exciting

MatheiBoulomenos

They are very different. I think of category theory more as a framework

Leaky Nun

@MatheiBoulomenos they are both very amazing and exciting to me

Daminark

I find the niftiness of category theory to stem from how you're able to put so many things in that context. It was built for the needs of AT, since you had to pass back and forth from topological spaces and homotopy classes of maps to groups and homomorphisms

And then it continued to evolve, for example with the whole Quillen model category business in which you can do homotopy theory, etc. But I've found it to be, while very fun, more of a language in which you can talk about other things, as opposed to an object of study

This is, in part, because it's so hard to make facts about all such structures, like the only theorem of category theory that I know of is Yoneda's lemma

I've heard the thing about Galois theory is more that it just has a really nice theory to it

MatheiBoulomenos

Yeah, that sounds about right

adjoint functors and their properties are very nifty as well

can save you a lot of computations

Kevin Driscoll

17:27

It is interesting how this sort of thing works. For example in physics, Newton's formulation and Hamilton's formulation of classical mechanics are totally equivalent. And passing from one to the other can be fruitful practically; some problems are just easier to solve one way than the other. But then also it can reveal some structure to the problems that you were unaware of previously.

In this case, the symplectic geometry

Daminark

17:41

Oh I remember symplectic geometry having something to do with physics, via Hamilton vector fields?

Actually I'm not sure if you know but is there some qualitative/philosophical reason why that should connect?

Like, I guess I don't see why having a manifold with a symplectic 2-form ought to be the structure that best frames mechanics

@Semi do you have any input on this?

Semiclassical

read Arnold, that's my input :P

Daminark

O lawd

Semiclassical

I don't really have a philosophical position on this, tbh

I think that if one wanted such a foundation, the best one classically would be Lagrangian mechanics

since that's got the principle of least action at its center

Daminark

I see

Secret

18:05

in The h Bar, 1 min ago, by Secret

> O and btw, don't bother pinging either Semiclassical nor Acuriousmind right now. They are so busy in philosophy that they are in what I called a lock-up state. Any group of users who are in lock-up state (typical example being balarka-Mike Miller), they usually won't respond to any pings that are "off topic"

I gauge the interesting level of a chat based on the number of lockup-states I can count at given moment in time

A chat level interesting level is anticorrelated to the number of lockup-states

So, the chat yesterday are full of lockup states related to Barlarka, Liad and other green gravatars

~~which is why it is boring enough for me to ponder about dedekind sunsets~~

Actually, not correct, the chat is most boring when there are no active users

Daniel Castle

Hi

Secret

as yesterday I am not even chatting much in the main chat

Kevin Driscoll

@Daminark Sorry, I don't feel like I understand symplectic geometry enough to give a reasonable opinion

Daniel Castle

The sequence 1,3,5,8,9,12,13,14,17,19,20,27,30,32,33,41,42,44... is of the remaining numbers in an adapted version of the sieve of eratosthenes, where instead of repeatedly eliminating numbers starting from n by adding n each time, you eliminate numbers by adding n+1 each time. Would anyone be able to help me reason about a proof of whether or not there are infinitely many of these numbers? Or perhaps derive some other useful properties about them

since the gap between each elimination diverges to +infinity, you would expect there to be a higher proportion of these numbers than primes

Mary Star

Hello!!

The mapping $X:M\rightarrow \mathcal{X}$ describes the sum of the two numbers rolled of two dice.
I want to determine the set $M$ and $X^{-1}(\{2,3,4\})$.

I have done the following:

Since the mapping describes the sum of the two numbers of the dice, the set $M$ contains every possible 2-tuple which describes every pair of numbers that we can get by the two dice.
So $M=\{1,2,3,4,5,6\}\times \{1,2,3,4,5,6\}$, right?

We have that
$$X^{-1}(\{2,3,4\})=\{(i,j)\in M \mid X(i,j) \in \{2,3,4\}\} =\{\text{every possible 2-tuple so that the sum is equal to 2,3 or 4}\}=\{(1,1), (1,2), (2,1), (1,3), …

Eric Silva

18:25

@Daminark see any of the good talks?

Kevin Driscoll

@MaryStar Sounds correct to me

Mary Star

@KevinDriscoll Great!! Thank you!! :-)

♦ $Mathematics$ Logic

Transcript for

$Mathematics$ Mathematics

♦ Logic

Transcript for

♦ $Mathematics$ Logic

$Mathematics$ Mathematics